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A polariser and an analyser are oriented so that maximum light is transmitted of intensity $I_0$ What will be the intensity of the out coming light when analyer is rotated through $60^{\circ}$

$(a)\;\frac{I_0}{2} \\ (b)\;\frac{I_0}{4} \\ (c)\;\frac{\sqrt 3 I_0 }{2} \\ (d)\;\frac{3 I_0}{4} $

1 Answer

According to Mauls law
$I= I_0 \cos^2 \theta$
$\quad= I_0 \cos ^2 60$
$\quad= I_0 \bigg[ \large\frac{1}{2}\bigg]^2$
$\quad= \large\frac{I_0}{4}$
Hence b is the correct answer.
answered Jan 23, 2014 by meena.p

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